HDOJ-1004

Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 103000    Accepted Submission(s): 39524

Problem Description

Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color and find the result.

This year, they decide to leave this lovely job to you. 

 

 

Input

Input contains multiple test cases. Each test case starts with a number N (0 < N <= 1000) -- the total number of balloons distributed. The next N lines contain one color each. The color of a balloon is a string of up to 15 lower-case letters.

A test case with N = 0 terminates the input and this test case is not to be processed.

 

 

Output

For each case, print the color of balloon for the most popular problem on a single line. It is guaranteed that there is a unique solution for each test case.

 

 

Sample Input

5

green

red

blue

red

red

3

pink

orange

pink

0

 

 

Sample Output

red

pink

 

本题题意：首先输入数据个数n，以下n行输入n个字符串表示气球颜色，输出出现次数最多的颜色。

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm> 
using namespace std;

const int MAX=1010;
int a[MAX];
char s[MAX][16];//本想用于最多15种颜色开的二维数组，后来发现一个问题 
int n;

int main(){
    while(scanf("%d",&n)&&n!=0){
        for(int i=0;i<n;i++){//输入 
            scanf("%s",&s[i]);
        }
        memset(a,0,sizeof(a));
        int ans=0;
        for(int i=0;i<n-1;i++){
            for(int j=i+1;j<n;j++){
            if(strcmp(s[i],s[j])==0){//遇到相等的气球则a[i]++ 
                a[i]++;
            }
        }
        if(a[i]>ans)
        ans=a[i];//ans为最大 
        }
        for(int i=0;i<n;i++){
            if(a[i]==ans)
            puts(s[i]);
        }
    }
    return 0;
}

PS：

代码注释中提到的问题，s数组中的[16]看起来并没有用上，可去掉后就会出现编译问题，求大神指教[16]在这里到底做了什么。